6.2.1: Illustrations

Illustration 1: Circuit Builder

\(\color{red}{\text{The JavaScript version of Circuit Builder is not yet fully functional. For example, the oscilloscope is not implemented.}}\)

Circuit Builder can be used to build and analyze DC and AC electrical circuits. Circuit Builder was written by Toon Van Hoecke at the Universiteit Gent.

If you want to change the size of the circuit grid, change the number of rows and columns and then press the "Set grid" button. Be aware that all components on the screen will disappear. The "Show ->" button is used to visualize arrows representing the chosen current direction. New components for the circuit are entered by a drag-and-drop method. Press the button of the component you want to add, hold the mouse button, and release it on the position where you want to put this component. Once you drop the component on the circuit grid, a box will appear that will allow you to change the parameters of the component. Any horizontal or vertical position between two black dots can be taken.

The following direction-independent (no positive or negative side) components can be added:

•  Resistor: value entered in ohms.
•  Capacitor: value entered in farads.
•  Inductor: value entered in henries.
•  Lightbulb: characterized by voltage (in \(\text{V}\)) and power (in \(\text{W}\)). The color varies from black (no current) to white (maximum current).
•  Wire: used to close connections.
•  Switch: can be opened and closed.

Some components are polarized and thus have a positive and a negative side. The direction can be set by choosing "\(+\) down/right" or "\(+\) up/left" on the Direction list item. The following direction-dependent components can be added:

•  Battery: value entered in \(\text{V}\).
•  General Voltage source: Its function prescription (in \(\text{V}\)) can be entered next to the source button and its default value is \(\sin (t\ast 2\ast \text{pi}\ast f\)). Use \(t\) as the time variable, \(f\) as the frequency variable, and "\(p\)" as the period variable.
•  Current Source: value entered in \(\text{A}\).
•  Oscilloscope: simulation of a one-beam oscilloscope. A window with the view on the particular oscilloscope can be opened by selecting the "Display Oscilloscope" option on the popup menu that appears when you click the right mouse button on the oscilloscope icon in the circuit.
•  Voltmeter: simulation of a digital voltmeter. Use the "Display Voltmeter" option on the right-mouse-button popup menu.
•  Ampèremeter: simulation of a digital ampèremeter. Use the "Display Ampèremeter" option on the right-mouse-button popup menu.

The "Calculate" button is used for recalculating data. A number of Step # data points is calculated iteratively with a step size of Step (\(\text{s}\)). The step size is entered in seconds (default is \(1\text{e}-6\text{ s}\)).

The "Start/Pause" button and the "Reset" button are used when a real-time clock is necessary. This is in situations with slow varying sources or the displaying of voltage or current graphs. The number of frames per second is \(1/(10\ast\text{step size})\). This is in real time up to step sizes of \(0.01\text{ s}\).

You can move a component to another position by using drag and drop. Other actions are available as options on the popup menu that appears when you click the right mouse button on the component's icon in the circuit. The possible options are only enabled if they are relevant:

• Delete Component: deletes the selected component.
• Change Value: changes the value or function of the selected component.
• Display Value Knob: pops up a little window with a scroll bar to change the value dynamically (linear steps or logarithmic steps).
• Display Frequency Knob: pops up a little window with a scroll bar to change the frequency of an AC source dynamically. This only works when the \(f\) variable is present in the function prescription.
• Show/Hide Value or Function: concerns the display on the circuit grid.
• Set Label: gives a name to the selected component.
• Display Oscilloscope: pops up the oscilloscope window of the selected oscilloscope.
• Display Voltmeter: pops up the digital voltmeter window of the selected voltmeter. The mode can be switched between DC and AC (rms value).
• Display Ampèremeter: pops up the digital ampèremeter window of the selected ampèremeter. The mode can be switched between DC and AC (rms value).
• Display Voltage Graph: pops up a voltage graph for the selected component (use "Start" button).
• Display Current Graph: pops up a voltage graph for the selected component (use "Start" button).
• Change Switch: changes the status of the selected switch, open or closed.
• Change Polarity: switches the \(+\) and \(-\) signs of components that are polarized.

Illustration authored by Toon Van Hoecke.
Applet authored by Toon Van Hoecke.

Illustration 2: AC Voltage and Current

Assume an ideal power supply. The graph shows the voltage \(\color{red}{\text{(red)}}\) across and current \(\color{blue}{\text{(blue)}}\) from the power supply as a function of time. Note the \(10^{-3}\) exponent on the timescale when the animation starts (voltage is given in volts, current is given in hundredths of amperes, and time is given in seconds). Restart.

Start the low voltage animation. As you change the frequency, describe what happens to the bulb and the graph. Note the factor of \(10^{-3}\) for the timescale when the animation starts. As you close the switch, notice that the voltage does not change, but the current increases. This is because closing the switch adds more resistors in parallel to the power supply.

Since the voltage is positive as much as it is negative, we do not talk about the average voltage (which would be zero) but instead describe the voltage either by the amplitude (the size of the peak voltage. What is it in this case?) or the rms (root-mean-square) voltage (\(= V_{\text{peak}}/\sqrt{2}\)). For this power supply, the peak voltage is \(5\text{ V}\) and the rms voltage is \(3.5\text{ V}\).

Household voltage is \(120\text{ V}\) rms. What is the peak voltage? In order to plot the current on the same graph, the current shown is \(100\) times the actual current. What is the average power of one lightbulb? (\(P = I_{\text{rms}}V_{\text{rms}}= V_{\text{peak}}I_{\text{peak}}/2\) for a purely resistive load). You should find that these lightbulbs are \(60\text{ W}\) bulbs.

With alternating current (AC), fluorescent lights in your room are flickering on and off \(120\) times/second (frequency in the US is \(60\text{ Hz}\)), but you simply don't notice it (just like a movie is made of separate frames, but to you it looks continuous). In Europe the standard frequency is \(50\text{ Hz}\), so the fluorescent lights in Europe go on and off \(100\) times in one second.

Illustration authored by Anne J. Cox.
Script authored by Wolfgang Christian and Anne J. Cox.

Illustration 3: Transformers

A transformer is connected to an outlet. The graph shows the input voltage (the voltage across the primary) and output voltage (the voltage across the secondary) as a function of time. Restart.

A transformer works by induction. A changing voltage in the primary coil (connected to the outlet) causes a changing current in the primary coil. The changing magnetic flux in the primary coil induces an emf (voltage) in the secondary coil. If you think about a coil of wire, the induced emf depends on the rate of change of the magnetic flux through the coil and the number of windings in the coil. Try changing the number of windings on the primary and secondary. How does the ratio of peak voltages depend on the windings? You should find that the ratio of the voltages is equal to the ratio of the coils. If the number of windings on the primary is greater than on the secondary, it is called a step-down transformer, but if the number of windings on the primary is smaller than on the secondary, it is a step-up transformer.

Both the step-up and step-down transformers conserve energy. For ideal transformers (no heat losses), energy conservation means that the average power (\(I_{\text{rms}}V_{\text{rms}}\)) is the same in the primary and secondary. Since the ratio of the number of windings is equal to the ratio of the voltages, for a step-down transformer with \(200\) turns on the primary and \(20\) turns on the secondary, \(2\text{ A}\) coming into the transformer would yield \(20\text{ A}\) out of the secondary. Conversely, for a step-up transformer, less current would be available at the secondary.

The facts that the power is the same in the primary and secondary and that transformers are easy to construct (coils wound around iron cores) are the reasons we use alternating current (instead of DC). Power companies can deliver a large amount of power either with high voltages and low current or with low voltages and high current. For the same amount of power, the lower-current option is preferable because of resistive heat losses on power lines. Consider the following two ways to deliver power over a \(10-\Omega\) power line and notice that the power from the plant is the same in both cases.

1. \(V = 10,000\text{ V}\) at the power plant and \(2\text{ A}\) through the line. The total power dissipated is given by \(I^{2}R = 40\text{ W}\) (and the voltage drop between the power plant and the user is \(20\text{ V}\)).
2. \(V = 1000\text{ V}\) at the power plant and \(20\text{ A}\) through the line. In comparison, the total power dissipated is 4000 W (and the voltage drop is 200 V).

It is clearly better to choose the high voltage, low current route, and so power plants produce electricity at high voltages (around \(20\text{ kV}\)). This is stepped up with transformers to a couple of hundred \(\text{kV}\) (e.g., \(300,000\text{ V}\)) for cross-country transmission and then stepped back down in cities and at your house. It is not nearly as easy or as efficient to step up and down DC, which makes AC cheaper to transport over wires.

Illustration authored by Anne J. Cox.
Script authored by Morten Brydensholt and modified by Anne J. Cox.

Illustration 4: Phase Shifts

Assume an ideal power supply. The graph shows the voltage \(\color{red}{\text{(red)}}\) across and current \(\color{blue}{\text{(blue)}}\) from the power supply as a function of time (voltage is given in volts, current is given in milliamperes, and time is given in seconds). Restart.

We start by reviewing the current and voltage relationship for a resistive load. As you change the frequency in the animation, what happens (if anything) to the ratio of voltage to current? Notice that the voltage and current are in phase with each other in this circuit.

Try a capacitive load. What happens to the amplitude of the current as you increase the frequency? The ratio of \(V/I\) is not called a resistance for this type of load; it is called the reactance (or impedance, but impedance includes information about the phase shift between voltage and current). This means that the reactance of a capacitive load changes with frequency. Since the current increases as the frequency increases, the reactance must decrease as the frequency increases.

Notice the phase shift between the current and the voltage. Pause the graph. Which plot (the voltage or the current) is in the "lead?" In other words, if you look at a time in which the current reaches its maximum value, has the voltage already reached its maximum value or will it reach its maximum value at a slightly later time? If the current reaches its maximum first, we describe this as the "current leading the voltage, " but if the voltage reaches its maximum first, we call it "current lagging the voltage." Which is the case with a capacitor?

Try an inductive load. What happens to the amplitude of the current as you increase the frequency? Does the current lead or lag the voltage in this case? Notice that with a capacitive load the current leads the voltage, while with an inductive load the current lags the voltage.

Because the current and voltage are out of phase when there are capacitive and inductive loads, and the reactance is a function of frequency, the mathematics to calculate the voltage and current is a bit more involved, but Kirchhoff's laws still hold at any instant in time.

Illustration authored by Anne J. Cox.
Script authored by Wolfgang Christian and Anne J. Cox.

Illustration 5: Power and Reactance

Assume an ideal power supply. The graph shows the voltage \(\color{red}{\text{(red)}}\) across the source and the current \(\text{(black)}\) through the circuit as a function of time (voltage is given in volts, current is given in milliamperes, and time is given in seconds). Restart.

Resistive circuit: Look at the plot of voltage and current. Power is given by \(P = VI\), but the current and voltage vary in time. It is more useful to think about the average power, which is \(P = V_{\text{rms}}I_{\text{rms}} = I_{\text{rms}}^{2}R = V_{\text{rms}}^{2}/R\). Notice that the current and voltage are always in phase, and so the product \(VI\) is always positive.

Capacitive circuit: Look at the plot of voltage and current. Notice that when the voltage is going from \(0\) to a more positive number, the current is going from a maximum value toward \(0\), but then when the voltage is going from its max back towards \(0\), the current has changed direction and is going from \(0\) down to a negative value. Thus, on the average over many cycles, the current and voltage are out of phase by \(\pi /2 = 90^{\circ}\). When the voltage is positive, the current is negative as much as it is positive, and the same applies when the voltage is negative. This means that the average power is \(0\). Compare this with the resistive load. When the voltage is positive, the current is positive, and when the voltage is negative, the current is negative. The resistor is always drawing current away from the source. Another way to think about this is that the capacitor simply stores charge. As the voltage changes direction, the current goes back and forth between the source and the capacitor, so the capacitor does not dissipate any energy over time (it simply stores the energy briefly).

Inductive circuit: Compare the plot of voltage and current for the inductor to that of the capacitor. Can you explain why the average power is \(0\) for this inductor just as it is for the capacitor?

If you have a circuit with a combination of resistive, capacitive, and inductive loads, calculating the average power dissipated now requires calculating \(V_{\text{rms}}I_{\text{rms}}\cos\phi\), where \(\phi\) is the phase shift between the current and the voltage (see Exploration 31.4).

Illustration authored by Anne J. Cox.
Script authored by Wolfgang Christian and Anne J. Cox.

Illustration 6: Voltage and Current Phasors

Assume ideal components. The bottom graph shows the voltage as a function of time across the source \(\color{red}{\text{(red)}}\), the resistor \(\color{blue}{\text{(blue)}}\), and the capacitor \(\color{green}{\text{(green)}}\). Current is shown in \(\text{black}\) (voltage is given in volts, current is given in milliamperes, and time is given in seconds). Restart.

You cannot simply use \(V = I R\) when working with AC circuits, because you must account for the phase differences in the voltages and currents. Notice that when you look at Voltages Only, the voltages across the power supply, the resistor, and the capacitor are not in phase. One way to account for the phase differences is to describe the voltage with phasors as shown in the animation in the top right box. The voltage of each component is represented by a vector that rotates at the frequency of the source. The angle between the vectors represents the phase difference between the voltages, while the length of the vectors represents the peak voltage across each circuit element. Illustration 31.7 and Explorations 31.5 and 31.6 develop this idea further. We can also use this phasor representation to describe the current. Look at Voltage and Current and notice that the voltage from the source is out of phase with the current. So, instead of using \(V = I R\), Ohm's Law becomes \(V = I Z\), where \(Z\) is the impedance and includes the frequency response and phase shift associated with the various circuit components.

Illustration authored by Anne J. Cox.
Script authored by Wolfgang Christian and Anne J. Cox.

Illustration 7: RC Circuits and Phasors

Assume an ideal power supply. The top graph shows the voltage as a function of time across the source \(\color{red}{\text{(red)}}\), the resistor \(\color{blue}{\text{(blue)}}\), and the capacitor \(\color{green}{\text{(green)}}\) (voltage is given in volts and time is given in seconds). Restart.

In order to analyze circuits with impedances that change as a function of frequency, we can use a phasor representation for the voltages across and the current through the various circuit elements. This allows us to take into account the phase difference between the voltages across the capacitor, the resistor, and the power supply.

To begin analyzing the circuit in the animation, we should first notice that Kirchhoff's law holds at any instant of time. Pause the animation and pick a time and find the voltage across the source, the resistor, and the capacitor. Verify that the voltage across the resistor plus the voltage across the capacitor equals the source voltage at that time. Notice, however, that if you add up the peak voltages, the peak resistor voltage plus the peak capacitor voltage is not equal to the peak source voltage.

You must account for the phase difference in the voltages. One way to account for the phase differences is to describe the voltage and current with phasors. Below the circuit are an animation and graph that show the phasor representation of the circuit elements (this allows us to show the phase difference), where the capacitor voltage is \(\pi /2\) behind the resistor since its voltage lags the current. Notice that the phasors rotate at an angular speed of \(\omega = 2\pi f\). The projection of the phasor vector on the \(y\) axis is plotted in the lower graph. Pause the animation. Note that the phasor animation matches the circuit graph. Try another frequency and verify that the two plots are the same (except at \(t = 0\) because of the initial condition of the capacitor). Therefore, we can use phasor diagrams to show the phase angle between the source voltage, the resistor voltage, and the capacitor voltage. For more on phasors, see Illustration 31.6 and Explorations 31.5 and 31.6.

Illustration authored by Anne J. Cox.
Script authored by Wolfgang Christian and Anne J. Cox.

Illustration 8: Impedance and Resonance, RLC Circuit

The impedance of a circuit is the relationship between the voltage and the current, \(V = I Z\), where \(Z\) is the impedance. In a purely resistive circuit, \(Z = R\) and the voltage and current are in phase. When capacitors and inductors are included, the relationship between voltage and current is more complicated. Calculating the impedance means taking the phase shift between voltage and current into account. In a series \(RLC\) circuit, the impedance is given by

\[Z=(R^{2}+(\omega L-1/\omega C)^{2})^{1/2}\nonumber\]

Notice that the impedance will be the smallest when \(\omega L = 1/\omega C\). If the impedance is smallest, what does that mean for the current at a given voltage? The frequency associated with this condition is called the resonant frequency. In the graph above you can see how the impedance changes as a function of frequency for different values of the resistance, the capacitance, and the inductance. If you change \(R\), does the resonant frequency change? What if you increase \(C\)? What if you increase \(L\)? Using the equation, you should be able to predict what will happen as you change the values. At resonance, the power dissipated is the greatest and the impedance is said to be purely ohmic because \(Z = R\) (since \(\omega L = 1/\omega C\)). Restart.

Illustration authored by Anne J. Cox.
Script authored by Morten Brydensholt.
Applet authored by Harry Broeders and modified by Wolfgang Christian.

Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.